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Recall. A commutative ring is a ring in which the multiplication operation is commutative,i.e., ∀a, b ∈ R, a·b = b·a.
Definition. Let R be a commutative ring. A polynomial ring is a ring formed from the set of polynomials in one or more indeterminates or variables with coefficients in R. R[x] = {anxn + an-1xn-1+ ··· + a1x + a0 | ai ∈ R, n ∈ ℤ+}, e.g. 2x3+ 7x2 +4∈ ℤ[x], 2⁄3x2 + 4x -7 ∈ ℚ[x].
The degree of f(x) = anxn + an-1xn-1+ ··· + a1x + a0, expressed or written as deg(f(x)), is the largest k such that the coefficient of ak is not zero. If an ≠ 0, we say that f(x) has degree n and its leading coefficient is an. A constant polynomial is either the zero polynomial or a polynomial of degree zero.
Two polynomials anxn + an-1xn-1+ ··· + a1x + a0 and bmxm + bm-1xm-1+ ··· + b1x + b0 are equal when the corresponding coefficients of xk are equal, ai = bi ∀i ∈ ℤ+, ai = 0 when m < i ≤ n, bj = 0 when n < j ≤ m
Let R be a commutative ring, and let f(x) = anxn + an-1xn-1+ ··· + a1x + a0 and g(x) = bmxm + bm-1xm-1+ ··· + b1x + b0 ∈ R[x]. The polynomial ring R[x] is equipped with addition and multiplication,
$(2x^3+x^2+2x+2)(2x^2+2x+1) = 4x^5+(4+2)x^4+(2+2+4)x^3+(1+4+4)x^2+(2+4)x+2=x^5+2x^3+2$
Theorem. Let R be a commutative ring with unity 1. R[x] is a commutative ring under the operations of polynomial addition and multiplication with unity (1, 0, 0, ···).
Proof.
It is left to the reader to verify that (R[x], +, ·) is a commutative ring with unity (1, 0, 0, · · ·). Besides, (0, 0, · · ·) is the zero element of R[x] and the addition inverse of (a0, a1, ···, an) is (-a0, -a1, ···, -an).
Futhermore, the mapping a → (a, 0, 0, ···) is a monomorphism (an injective homomorphism) of the ring R into R[x]. Thus (R ≋ Φ(R[x])), R can be considered as a subring of R[x] and we no longer need to distinguish between a and (a, 0, 0, ···).
Theorem. Let D be a commutative ring with unity 1. Then, D[x] is an integral domain if and only if D is an integral domain.
Proof.
Recall that an integral domain is a commutative ring with unity in which the product of any two nonzero elements is nonzero.
⇒ ) Suppose D[x] is an integral domain. D is a subring of D[x] (D ⊆ D[x]), then it is obvious that D must also be commutative ring with unity(🚀) and has no zero divisors (∀x, y ∈ D with x·y = 0, x, y ∈ D[x] ⇒[D[x] is an integral domain] x = 0 or y = 0).
🚀 Consider the map, Φ:R[x] → R, such that Φ(f(x)) = Φ(a0 + a1x + ··· + anxn) = a0. Φ is an onto homomorphism (it is left as an easy exercise) ⇒ R is a homomorphic image of R[x], a ring with unity and [A homomorphic image of a ring with unity is a ring with unity] Φ(e(x)) will be the unity of R where e(x) is the unity of R[x].
⇐ ) Conversely, suppose D be an integral domain.
f(x)·g(x) = cm+nxm+n + cm+n-1xm+n-1+ ··· + c1x + c0 where ck = akb0 + ak-1b1 + ··· + a1bk-1 + a0bk, for k = 0,···, m+n
Its leading coefficient is cm+n = anbm ≠ 0 because D is an integral domain. Therefore, f(x) ≠ 0 and g(x) ≠ 0 ⇒ f(x)·g(x) ≠ 0, hence D[x] has no zero-divisors.
Theorem. Let R be an integral domain, let f and g be two nonzero polynomials in R[x], then deg(fg) = deg(f) + deg(g).
Proof:
Suppose f(x) = anxn + an-1xn-1+ ··· + a1x + a0 and g(x) = bmxm + bm-1xm-1+ ··· + b1x + b0 are nonzero polynomials of degree n and m respectively, i.e., an, bm ≠ 0R ⇒ the highest possible degree of fg is n + m and the coefficient of xn+m is anbm.
Since R is an integral domain, anbm = 0R if and only if an = 0R or bm = 0R. Therefore, an, bm ≠ 0R ⇒ an·bm ≠ 0R ⇒ deg(fg) = n + m = deg(f) + deg(g).
Theorem. Let R be an integral domain, then the units of R[x] are the units of R.
Proof:
Suppose f(x) = anxn + an-1xn-1+ ··· + a1x + a0 is a unit ⇒ there exist g(x) ∈ R[x], say deg(f(x)) = n, and deg(g(x)) = m, such that f(x)·g(x) = 1 ⇒[deg(fg) = deg(f) + deg(g), f(x)·g(x) = 1] n + m = 0 ⇒ n = m = 0 ⇒ f(x) = u ∈ R, g(x) = v ∈ R, u·v = 1, and therefore u and v are units in R.
Theorem. If I is an ideal of a ring R, define I[x] = {i0 +i1x +··· + inxn| ij∈ I}. Then, R[x]/I[x] ≋ (R/I)[x]
Proof.
Consider Φ: R[x] → (R/I)[x] given by Φ(a0 +a1x +··· + anxn) = (a0 + I) +(a1 + I)x +··· + (an + I)xn
Φ is a surjective homomorphism.
∀f(x), g(x) ∈ R[x], f(x) = anxn + an-1xn-1+ ··· + a1x + a0 and g(x) = bmxm + bm-1xm-1+ ··· + b1x + b0
Φ(f(x) + g(x)) = Φ($\sum_{i=0}^{max(m,n)} (a_i+b_i)x^i$) = $\sum_{i=0}^{max(m,n)} (a_i+b_i+I)x^i$ =[By the definion of coset addition] $\sum_{i=0}^{max(m,n)} (a_i+I)x^i+\sum_{i=0}^{max(m,n)} (b_i+I)x^i$ = Φ(f(x)) + Φ(g(x))
Similarly, Φ(f(x) · g(x)) = Φ(f(x))·Φ(g(x)). Futhermore, Φ is onto. Let F[x] ∈ (R/I)[x], then F[x] = $\bar a_0 + \bar a_1x + ··· + \bar a_nx^n,\bar a_0 = a_0 + I, \bar a_1 = a_1 + I, ···, \bar a_n = a_n + I, a_0, a_1, ···, a_n ∈ R$. Set f(x) = a0 + a1x + ··· + anxn ∈ R[x], and obviously Φ(f(x)) = F[x] and F[x] was taken an arbitrary element of (R/I)[x].
f(x) = a0 + a1x + ··· + anxn ∈ Ker(Φ) ↭ ak + I = 0 + I ∀0 ≤ k ≤ n ↭ ak ∈ I ∀0 ≤ k ≤ n ↭ f(x) ∈ I[x], hence ker(Φ) = I[x]. Finally, by the First Isomorphism Theorem for rings, R[x]/I[x] ≋ (R/I)[x]
Theorem. P ⊆ R is a prime ideal if and only if P[x] ⊆ R[x] is a prime ideal.
Proof.
Recall: A prime ideal A of a commutative ring R is a proper ideal of R such that if a, b are two elements of R, and whenever their product ab is an element of A, then either a is in A, b is in A or both, i.e., a, b ∈ R, ab ∈ A ⇒ a ∈ A or b ∈ A. Example: ⟨x⟩ ⊆ ℤ[x] is a prime ideal, but ⟨x2⟩ is not, because x2 ∈ ⟨x2⟩ but x ∉ ⟨x2⟩.
P ⊆ R is a prime ideal ↭[Theorem. Let R be a commutative ring with unity, and let A be an ideal of R. Then, A is a prime ideal iff R/A is an integral domain.] R/P is an integral domain ↭[Theorem. Let D be a commutative ring with unity 1. Then, D[x] is an integral domain if and only if D is an integral domain.] (R/P)[x] is an integral domain ↭[Theorem. If I is an ideal of a ring R, define I[x] = {i0 +i1x +··· + inxn| ij∈ I}. Then, R[x]/I[x] ≋ (R/I)[x]] R[x]/P[x] is an integral domain ↭ P[x] is a prime ideal ∎
Division Algorithm. Let F be a field, f(x), g(x) ∈ F[x], g(x) ≠ 0. Then, there exists unique polynomials q(x) and r(x) ∈ F[x] such that
Proof.
We first prove the existence of the polynomials q and r. There are three possibilities:
3.1 If deg(f) = 0 ⇒ deg(g) = 0 ⇒ f = a, g = b, for some nonzero elements a and b, a, b ∈ F ⇒ [F is a field, and by assumption, b = g(x) ≠ 0, and therefore b is a unit, ∃b-1: bb-1 = b-1b = 1] f = a = b(b-1a) = g(b-1a), the theorem holds with q = b-1a, and r = 0F
3.2 Let’s assume that the proposition holds whenever deg(f) < n. We must show that it is true when f(x) has degree n, say f(x) = anxn + an-1xn-1+ ··· + a1x + a0 with an ≠ 0F. The divisor g(x) = bmxm + bm-1xm-1+ ··· + b1x + b0, bm ≠ 0F, and m ≤ n.
bm ≠ 0 (By assumption, g(x) ≠ 0), F is a field ⇒ bm is an unit. Next, we multiply the divisor g by anbm-1xn-m (n -m ≥ 0) to obtain (anbm-1xn-m·g) = anxn + anbm-1bm-1xn-1 + ··· + anbm-1b0xn-m
Since the leading term of this polynomial is identical to that of f, namely an, the difference of f - anbm-1xn-m·g is a polynomial of degree less than n, and we can apply the induction hypothesis with g as the divisor polynomial and f - anbm-1xn-m·g as the dividend ⇒ ∃ polynomials q1 and r:
f - anbm-1xn-m·g = q1g + r, r = 0F or deg(r) < deg(g).
f = (anbm-1xn-m + q1)·g + r, r = 0F or deg(r) < deg(g) ∎
To prove that q and r are unique, suppose that q’ and r’ are polynomials satisfying f = q’g + r’, r’ = 0F and deg(r’) < deg(g).
qg + r = f = q’g + r’ ⇒ g(q - q’) = r’ - r.
If q - q’ ≠ 0F ⇒ the degree of the polynomial on the right hand size of the last equation (r’ - r) is greater than or equal to the degree of g [deg(r’-r) ≥ deg(g) + deg(q -q’) ⇒ deg(r’-r) ≥ deg(g)]. But this is not possible since the polynomials r’ and r are either zero or have degree strictly less that of g ⇒ q’ = q ⇒[By assumption, g(x) ≠ 0] r’ = r ∎
Example: x4 -2x3 -x +1 = (x2 + x -1)(x2 +2x -1) + 2x in ℤ5[x] (Figure 1.b.). x4 -2x3 -x +1 = (x2 - 4x +9)(x2 +2x -1) + (-23x + 10) in ℤ (Figure 1.c.)